3 Answers
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HINT $\ \: $ Distributivity easily yields that a finitely generated ideal is $\:1\:$ if it contains a cancellable element $\rm\:u\:$ that is $\rm\:lcm$-coprime to the generators. For example, for a $2$-generated ideal $\rm\:(x,y)$

@Bill I have an irrelevant question which is nevertheless relevant to my interests. Why do you always use the \rm version of your math variables? Is it just a personal, stylistic choice?
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barfAug 18 '11 at 19:46

A commutative integral domain is "arithmetic" in the sense that you specify iff it is a Prüfer domain, i.e., iff every nonzero finitely generated ideal is invertible. This class of domains is famously robust: there is an incredibly long list of equivalent characterizations: see e.g. the beginning of this paper for some characterizations. For a proof that a domain is arithmetic iff its finitely generated ideals are invertible, see e.g. Theorem 6.6 of Larsen and McCarthy's text Multiplicative Theory of Ideals.

Note in particular that a Noetherian domain is Prüfer iff it is Dedekind, i.e., iff it is integrally closed and of Krull dimension at most one. Therefore examples of rings with non-distributive lattice of ideals abound, e.g.:

For any field $k$, $k[t_1,\ldots,t_n]$, $n \geq 2$. (The dimension is greater than one.)
For any nonfield Noetherian domain $k$, $k[t_1,\ldots,t_n]$, $n \geq 1$. (The dimension is greater than one.)
$\mathbb{Z}[\sqrt{-3}]$, $k[t^2,t^3]$ for any field $k$. (The rings are not integrally closed.)

+1. Much better answer! (I should add: much better than mine, in case there will be other answers...)
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Pierre-Yves GaillardAug 18 '11 at 12:00

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@Bill: The paper linked to in my answer cites references where the equivalences are proved. If you (or the OP, or anyone else) do not have access to any of these references but are still interested in the proof, probably someone here could help you out. (Sorry for not wanting to argue about whether something is "simple": surely it's best for you to leave your own answer, as you have.)
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Pete L. ClarkAug 18 '11 at 19:06

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@Bill: Actually I said that I do not want to argue about whether my answer is "simple" (or whether the word "simple" in the OP's answer is meant to apply to the ring itself or to the proof that it is not arithmetic, etc.). Again, I apologize. If none of the answers given so far are satisfactory to the OP, he will surely tell us so.
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Pete L. ClarkAug 18 '11 at 19:22

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What's the beef here? Surely there's a better way to resolve this...
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The Chaz 2.0Aug 18 '11 at 19:45

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@Gonzalo If you continue your study of these rings I highly recommend that you do study the many diverse characterizations. They are both important and fascinating (that's why I typed $27$ of them into my answer last Decemeber!) But they are a bit overkill for the simple question that you posed. Note: not all of the $27$ I mentioned are in the Bazzoni and Glaz paper that I cite in the comments there. There are also many other interesting characterizations in the literature. Perhaps more than for any other algebraic structure.
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Bill DubuqueAug 19 '11 at 1:20